BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 63RE

(a)

To determine

To find:

The beam of maximal cross sectional area is square.

Expert Solution

Answer to Problem 63RE

The beam of maximal cross sectional area is square.

  A=x4r2x2

Explanation of Solution

Given:

The radius of the cylinder

  r=10inches

Concept used:

Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.

Calculation:

Let x be depth of the rectangle and y be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length 2r .

By using Pythagoras theorem

  x2+y2=(2r)2y2=(2r)2x2y=4r2x2

The area of the rectangule A=xy................................(1)

Now after substituting 4r2x2 for y in equation (1)

  A=x4r2x2

(b)

To determine

To find:

The dimensions of the planks that will have maximal cross-sectional area.

Expert Solution

Answer to Problem 63RE

The dimensions of the planks that will have maximal cross-sectional area is x=2r .

Explanation of Solution

Given:

The radius of the cylinder

  r=10inches

Concept used:

Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.

Calculation:

Let x be depth of the rectangle and y be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length 2r .

By using Pythagoras theorem

  x2+y2=(2r)2y2=(2r)2x2y=4r2x2

The area of the rectangule A=xy................................(1)

Now after substituting 4r2x2 for y in equation (1)

  A=x4r2x2...............................(2)

Differentiating with respect to x

  dAdx=ddxx4r2x2dAdx=xddx4r2x2+4r2x2dxdxdAdx=4r2x22x24r2x2

Using the zero product property of the equation

  0=4r2x22x24r2x20=4r2x22x20=4r2x2x=±2rx=2r

(c)

To determine

To find:

The dimensions of the strongest beam that cut from the cylindrical log.

Expert Solution

Answer to Problem 63RE

The dimensions of the strongest beam that cut from the cylindrical log is x4r2x2=y

Explanation of Solution

Given:

The radius of the cylinder

  r=10inches

Concept used:

If the quantities x and y are related by an equation

  yαx

For some constant k0

That is y varies inversely proportional as x or y is directly proportional to x

The constant k is called constant proportionality

Calculation:

Let x be depth of the rectangle and y be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length 2r .

By using Pythagoras theorem

  x2+y2=(2r)2y2=(2r)2x2y=4r2x2

The area of the rectangule A=xy................................(1)

Now after substituting 4r2x2 for y in equation (1)

  A=x4r2x2

If the quantities x and y are related by an equation

  yαxy=kxx4r2x2=y

For some constant k0

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